How Long Does It Take a Bullet to Traverse a Wall with Variable Resistance?

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The discussion focuses on calculating the time it takes for a bullet to traverse a wall of thickness h, entering at speed v1 and exiting at speed v2, while experiencing a force proportional to the square of its velocity. The initial approach using Newton's second law leads to an expression for time that lacks dependence on wall thickness, prompting concerns about the correctness of the equation. It is suggested that the force should indeed depend on the position within the wall, linking the acceleration to the bullet's position. A differential equation is proposed to relate velocity and position, which can help derive a more accurate expression for time that incorporates the unknown ratio of mass to the constant. The discussion concludes with acknowledgment of the helpful insights provided.
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bullet enters a wall with speed v1 and goes out with speed v2, the wall is h units thick. During that time it experiences an impending force which is proprtional to v^2. I have to find the time the bullet goes out the wall.

I can take t0=0 v0=v1 x0=0

Using N2 law i get d(mv)/dt = -F.....mdv/dt = -m*v^2 (im not sure if I am missing here something)
dv/dt = -v^2

dt = - dv/v^2 ===> t = (v1-v2)/v1*v2...this result is weird since it doesn't contain h so i must have constructed the equation wrongly i guess.
I think that F depends on h since F can act only between [0; h], and maybe the answer should be multiplied with h and t = h*(v1-v2)/v1*v2...but it still doesn't look right to me...any opinions?
 
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vabamyyr said:
bullet enters a wall with speed v1 and goes out with speed v2, the wall is h units thick. During that time it experiences an impending force which is proprtional to v^2. I have to find the time the bullet goes out the wall.
I can take t0=0 v0=v1 x0=0
Using N2 law i get d(mv)/dt = -F.....mdv/dt = -m*v^2 (im not sure if I am missing here something)

Well, the force was said to be proportional to v^2.
So F= -kv^2, where k is a constant
and dv/dt=-k/m* V^2

[Quote}

dv/dt = -v^2
dt = - dv/v^2 ===> t = (v1-v2)/v1*v2...this result is weird ...[/QUOTE]

Including that constant k and m the mass of the bullet, you get

t=(v1-v2)/(v1*v2)*m/k (A)

But you do not know m/k. You should get it from h, the distance travelled.

I think that F depends on h since F can act only between [0; h]

You are on the right track. F depends on the position of the bullet inside the wall through the velocity. So the acceleration canbe considered as function of the position, x.

a = dv/dt = dv/dx * dx/dt = dv/dx *v.
F= ma = - kv^2

m*dv/dx *v = - k*v^2,

or dv/dx = (-k/m)*v

Solve this differential equation with the condition that v=v1 at x=0 and v=v2 at x=h. You get an expression containing the unknown k/m which you can use to determine t from equation (A).

ehild
 
thanks, i liked ur comments and u have been most helpful
 
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